Groups of the from $gMg$ in a monoid where $g$ is an idempotent Let $(M, \cdot)$ be a finite monoid with identity $e$. It is easy to see that $gMg = \{ gxg : x \in M \}$ forms a monoid with identity $geg = g$ if $g$ is an idempotent. If $gMg$ contains no idempotent other than $g$, it must be a group, since it is finite.
Suppose that $g, h \in M$ are distinct idempotents in $M$ such that both $gMg$ and $hMh$ are groups. Is it true that $gMg$ and $hMh$ are isomorphic?
 A: Yes. You do not even need the finiteness of $M$.

Theorem 1. Let $\left(  M,\cdot\right)  $ be a monoid with identity
  $e$. For every idempotent $g\in M$, the set $gMg=\left\{  gxg\mid x\in
M\right\}  $ is a monoid (with respect to the operation $\cdot$) with neutral
  element $geg=gg=g$. Let $g$ and $h$ be two idempotents of $M$ such that the
  monoids $gMg$ and $hMh$ are groups. Then, these groups $gMg$ and $hMh$ are isomorphic.

Let us first show the following simple fact:

Lemma 2. Let $\left(  A,\cdot\right)  $ and $\left(  C,\cdot\right)  $ be
  two groups. Let $\varphi:A\rightarrow C$ be a map. Assume that $\varphi\left(
a\right)  \varphi\left(  b\right)  =\varphi\left(  ab\right)  $ for all $a\in
A$ and $b\in A$. Then, $\varphi$ is a group homomorphism from $\left(
A,\cdot\right)  $ to $\left(  C,\cdot\right)  $.

Proof of Lemma 2. Let $e$ denote the neutral element of any group. We have
assumed that
\begin{equation}
\varphi\left(  a\right)  \varphi\left(  b\right)  =\varphi\left(
ab\right)  \qquad \text{for all $a\in A$ and $b\in A$.}
\label{darij.pf.l2.1}
\tag{1}
\end{equation}
Applying this to $a=e$ and $b=e$, we obtain $\varphi\left(  e\right)
\varphi\left(  e\right)  =\varphi\left(  \underbrace{ee}_{=e}\right)
=\varphi\left(  e\right)  $. Since $C$ is a group, we can cancel
$\varphi\left(  e\right)  $ from this equality, and obtain $\varphi\left(
e\right)  =e$. Combined with our assumption \eqref{darij.pf.l2.1}, this shows that $\varphi$
is a monoid homomorphism from $\left(  A,\cdot\right)  $ to $\left(
C,\cdot\right)  $. Thus, $\varphi$ is also a group homomorphism (since every
monoid homomorphism between groups is a group homomorphism). Lemma 2 is proven. $\blacksquare$
Proof of Theorem 1. The elements $g$ and $h$ are idempotents; thus, we have
$gg=g$ and $hh=h$.
Let $p$ be the inverse of the element $ghg$ in the group $gMg$. Then,
$p\left(  ghg\right)  =\left(  ghg\right)  p=g$ (since $g$ is the neutral
element of the group $gMg$).
Let $q$ be the inverse of the element $hgh$ in the group $hMh$. Then,
$q\left(  hgh\right)  =\left(  hgh\right)  q=h$ (since $h$ is the neutral
element of the group $hMh$).
The element $p$ belongs to the group $gMg$, but $g$ is the neutral element of
this group. Hence, $pg=gp=p$. Similarly, $qh=hq=q$.
We have
\begin{align}
gqhghgp=g\underbrace{q\left(  hgh\right)  }_{=h}gp=ghgp=\left(  ghg\right)
p=g ,
\end{align}
so that
\begin{align}
g
&= gqhghgp
= gqh \underbrace{\left(  ghg\right)  p}_{=g}
= g\underbrace{qh}_{=q}g
\label{darij.pf.thm1.2}
\tag{2}
\\
&= gqg .
\label{darij.pf.thm1.3}
\tag{3}
\end{align}
Similarly,
\begin{equation}
h = hph .
\label{darij.pf.thm1.4}
\tag{4}
\end{equation}
We define a map $\alpha:gMg\rightarrow hMh$ by setting
\begin{align}
\alpha\left(  x\right)  =hxq \qquad \text{ for every $x\in gMg$.}
\end{align}
This is well-defined, because every $x\in gMg$ satisfies $hx\underbrace{q}
_{\in hMh}\in h\underbrace{xhM}_{\subseteq M}h\subseteq hMh$.
We define a map $\beta:hMh\rightarrow gMg$ by setting
\begin{align}
\beta\left(  y\right)  =pyg \qquad \text{ for every $y\in hMh$.}
\end{align}
This is well-defined, because every $y\in hMh$ satisfies $\underbrace{p}_{\in
gMg}yg\in g\underbrace{Mgy}_{\subseteq M}g\subseteq gMg$.
Let us now prove that
\begin{align}
\alpha\left(  a\right)  \alpha\left(  b\right)  =\alpha\left(
ab\right)  \qquad \text{ for all $a\in gMg$ and $b\in gMg$.}
\label{darij.pf.thm1.5}
\tag{5}
\end{align}
[Proof of \eqref{darij.pf.thm1.5}: Let $a\in gMg$ and $b\in gMg$.
We have $a\in gMg$; thus, there exists some $x\in M$ such that $a=gxg$.
Consider this $x$. Thus, $\underbrace{a}_{=gxg}g=gx\underbrace{gg}_{=g}=gxg=a$.
We have $b\in gMg$; thus, there exists some $y\in M$ such that $b=gyg$.
Consider this $y$. Thus, $g\underbrace{b}_{=gyg}=\underbrace{gg}_{=g}yg=gyg=b$.
The definition of $\alpha$ yields $\alpha\left(  a\right)  =h\underbrace{a}
_{=ag}q=hagq$ and $\alpha\left(  b\right)  =h\underbrace{b}_{=gb}q=hgbq$.
Multiplying these two equalities, we obtain
\begin{align}
\alpha\left(  a\right)  \alpha\left(  b\right)  =ha\underbrace{gqhg}
_{\substack{=g\\\text{(by \eqref{darij.pf.thm1.2})}}}bq=h\underbrace{ag}_{=a}bq=habq .
\label{darij.pf.thm1.6}
\tag{6}
\end{align}
On the other hand, the definition of $\alpha$ yields $\alpha\left(  ab\right)
=habq$. Compared with \eqref{darij.pf.thm1.6}, this yields $\alpha\left(  a\right)
\alpha\left(  b\right)  =\alpha\left(  ab\right)  $. Thus, \eqref{darij.pf.thm1.5} is proven.]
Now, Lemma 2 (applied to $A=gMg$, $C=hMh$ and $\varphi=\alpha$) yields that
$\alpha$ is a group homomorphism from $\left(  gMg,\cdot\right)  $ to $\left(
hMh,\cdot\right)  $. We shall now focus on proving that $\alpha$ is invertible.
Indeed, let us first show that $\beta\circ\alpha=\operatorname*{id}$. Indeed,
let $x\in gMg$. Then, $xg=gx=x$ (since $x$ belongs to the group $gMg$, but $g$
is the neutral element of this group). We have
\begin{align}
\left(  \beta\circ\alpha\right)  \left(  x\right)
&=\beta\left(
\underbrace{\alpha\left(  x\right)  }_{=hxq}\right)  =\beta\left(  hxq\right) \\
&=\underbrace{p}_{=pg}h\underbrace{x}_{=xg}qg
\qquad \text{(by the definition of $\beta$)} \\
&=pgh\underbrace{x}_{=gx}\underbrace{gqg}_{\substack{=g\\\text{(by
\eqref{darij.pf.thm1.3})}}}=\underbrace{pghg}_{=p\left(  ghg\right)  =g}\underbrace{xg}
_{=x}=gx=x=\operatorname*{id}\left(  x\right) .
\end{align}
Thus, we have shown that $\left(  \beta\circ\alpha\right)  \left(  x\right)
=\operatorname*{id}\left(  x\right)  $ for every $x\in gMg$. This proves
$\beta\circ\alpha=\operatorname*{id}$.
Now, let us show that $\alpha\circ\beta=\operatorname*{id}$. Indeed, let $y\in
hMh$. Then, $hy=yh=y$ (since $y$ belongs to the group $hMh$, but $h$ is the
neutral element of this group). We have
\begin{align}
\left(  \alpha\circ\beta\right)  \left(  y\right)
&=\alpha\left(
\underbrace{\beta\left(  y\right)  }_{=pyg}\right)  =\alpha\left(  pyg\right) \\
&= hp\underbrace{y}_{=hy}g\underbrace{q}_{=hq}
\qquad \text{(by the definition of $\alpha$)} \\
&=\underbrace{hph}_{\substack{=h\\\text{(by \eqref{darij.pf.thm1.4})}}}\underbrace{y}
_{=yh}ghq=\underbrace{hy}_{=y}\underbrace{hghq}_{=\left(  hgh\right)
q=h}=yh=y=\operatorname*{id}\left(  y\right) .
\end{align}
Thus, we have shown that $\left(  \alpha\circ\beta\right)  \left(  y\right)
=\operatorname*{id}\left(  y\right)  $ for every $y\in hMh$. This proves
$\alpha\circ\beta=\operatorname*{id}$.
The maps $\alpha$ and $\beta$ are mutually inverse (since $\alpha\circ
\beta=\operatorname*{id}$ and $\beta\circ\alpha=\operatorname*{id}$). Thus,
the map $\alpha$ is invertible. Since $\alpha$ is a group homomorphism, this
shows that $\alpha$ is a group isomorphism. Thus, there exists a group
isomorphism $gMg\rightarrow hMh$ (namely, $\alpha$). This proves Theorem 1. $\blacksquare$
This proof was obtained by lots of experimentation and iterative
simplification. Do not ask me for the intuition behind it, for I have none. I
suspect it could still be shortened twice, but I am happy enough that I was
able to dispose of the finiteness condition and at least some of the ugliest
computations. Feel free to improve!
A: This is true for any semigroup (even if it is not a monoid) and is easy to prove if you know about Green's relations.
If $S$ be a semigroup and $e$ is an idempotent of $S$, then $eSe$ is a monoid
(with identity $e = eee$). Suppose that $e$ and $f$ are idempotents such that $eSe$ and $fSf$ are groups. Then in particular, $e \mathrel{\mathcal H} efe$ and $f \mathrel{\mathcal H} fef$, which implies that $e \mathrel{\mathcal R} ef$ and 
$ef \mathrel{\mathcal L} f$, whence $e \mathrel{\mathcal D} f$. It follows by Green's Lemma, than the two groups $eSe$ and $fSf$, which are maximal groups of a regular $\mathcal D$-class, are isomorphic.
